Should be 1.

It doesnt matter what the value of x is..

The way number 7 works is that the unit's digit repeats after 7^4.. so everytime you will get a sequence of 7,9,3,1.. so

I can write the equation as 7^(12x+3) = 7^12x*7^3

so 7^12x will always have units digit of 1.. when multiplied with 7^3, the unit's digit will always be 3.. now if you add 3 to the number.. the unit's digit will become 6..

So when you divide any number whose unit's digit is 6 by 5, the remainder has to be 1.

7^(12x+3), x>0
=>7^15, 7^27, 7^39 and so on

7^1, 7^2, 7^3, 7^4 have units digits as 7, 9, 3, 1 respectively. the pattern repeats every 4th power.

7^15, 7^27, 7^39 ....all will have the units digit as 3 (as 15, 27, 39..all leave a remainder of 3 when divided by 4)

last digit of the sum of the numerator: 6. will always leave a remainder of 1 when divided by 5

hence, B

The key is to deal with the last digit only. I get it!
Brilliant guys! Thanks!